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The Birman-Schwinger principle says that if $\Delta$ is the usual Laplacian on $\mathbb{R}^n$ and we consider the operator $H=-\Delta-V$ for a positive potential $V$, then, for any $\lambda>0$, the number of eigenvalues at most $-\lambda$ of this operator is the same as the number of eigenvalues at least 1 of the operator

$$ K_\lambda=V^{1/2}(-\Delta+\lambda)^{-1}V^{1/2}.$$

Can you suggest any source where this result is proved for $\lambda=0$?

Many thanks!

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  • $\begingroup$ What is $(-\Delta)^{-1}$? $\endgroup$
    – Fedor Petrov
    Dec 13, 2016 at 21:15
  • $\begingroup$ We can define it for example via $\mathcal{F}((-\Delta)^{-1}f)(x)=|x|^{-2}\mathcal{F}(f)(x)$, where $\mathcal{F}$ is just Fourier transform. $\endgroup$
    – cav
    Dec 13, 2016 at 23:33

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This runs into obvious technical issues. For example, $K_0$ will not be bounded (let alone compact) even for very nice $V$. So one also has to think about what exactly one wants to prove.

This paper discusses these issues. In particular, the Birman-Schwinger principle for $\lambda=0$ is stated in equation (1.6).

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